CBSE Class X

Question 2:

Construct a triangle of sides 4 cm, 5cm and 6cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle. Give the justification of the construction.

Answer:

Step 1

Draw a line segment AB = 4 cm. Taking point A as centre, draw an arc of 5 cm radius. Similarly, taking point B as its centre, draw an arc of 6 cm radius. These arcs will intersect each other at point C. Now, AC = 5 cm and BC = 6 cm and ΔABC is the required triangle.

Step 2

Draw a ray AX making an acute angle with line AB on the opposite side of vertex C.

Step 3

Locate 3 points A₁, A₂, A₃ (as 3 is greater between 2 and 3) on line AX such that AA₁ = A₁A₂ = A₂A₃.

Step 4

Join BA₃ and draw a line through A₂ parallel to BA₃ to intersect AB at point B’.

Step 5

Draw a line through B’ parallel to the line BC to intersect AC at C’.

ΔAB’C’ is the required triangle.

 

Justification

The construction can be justified by proving that

By construction, we have B’C’ || BC

∴ ∠AB’C’ = ∠ABC (Corresponding angles)

In ΔAB’C’ and ΔABC,

∠AB’C’ = ∠ABC (Proved above)

∠B’AC’ = ∠BAC (Common)

∴ ΔAB’C’ ∼ ΔABC (AA similarity criterion)

 ⇒ AB’/AB = B’C’/BC = AC’/AC ……….  (1)

In ΔAA₂B’ and ΔAA₃B,

∠A₂AB’ = ∠A₃AB (Common)

∠AA₂B’ = ∠AA₃B (Corresponding angles)

∴ ΔAA₂B’ ∼ ΔAA₃B (AA similarity criterion)

⇒ AB’/AB = AA₂/AA₃

⇒ AB’/AB = 2/3  ………………..  (2)

From equations (1) and (2), we obtain

AB’/AB = B’C’/BC = AC’/AC = 2/3

⇒ AB’ = 2/3 AB, B’C’ = 2/3 BC , AC’ = 2/3 AC

This justifies the construction.

Question 1:

Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the twoparts. Give the justification of the construction.

Answer:

A line segment of length 7.6 cm can be divided in the ratio of 5:8 as follows.

Step 1 Draw line segment AB of 7.6 cm and draw a ray AX making an acute angle with line segment AB.

Step 2 Locate 13 (= 5 + 8) points, A₁, A₂, A₃, A₄ …….. A₁₃, on AX such that AA₁ = A₁A₂ = A₂A₃ and so on.

Step 3 Join BA₁₃.

Step 4 Through the point A5, draw a line parallel to BA₁₃ (by making an angle equal to ∠AA₁₃B) at A₅ intersecting AB at point C.

C is the point dividing line segment AB of 7.6 cm in the required ratio of 5:8.

The lengths of AC and CB can be measured. It comes out to 2.9 cm and 4.7 cm respectively.

Justification

The construction can be justified by proving thata

AC/CB = 5/8

By construction, we have A₅C || A₁₃B. By applying Basic proportionality theorem for the triangle AA₁₃B, we obtain

AC/CB = AA₅/A₅A₁₃ ….   (1)

From the figure, it can be observed that AA₅ and A₅A₁₃ contain 5 and 8 equal divisions of line segments respectively.

∴ AA₅/A₅A_{13 = 5/8 …..  (2)}

On comparing equations (1) and (2), we obtain

AC/CB = 5/8

This justifies the construction.

Question 13:

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Answer:

Let ABCD be a quadrilateral circumscribing a circle centered at O such that it touches the circle at point P, Q, R, S. Let us join the vertices of the quadrilateral ABCD to the center of the circle.

Consider ΔOAP and ΔOAS,

AP = AS (Tangents from the same point)

OP = OS (Radii of the same circle)

OA = OA (Common side)

ΔOAP ≅ ΔOAS (SSS congruence criterion)

Therefore, A ↔ A, P ↔ S, O ↔ O

And thus, ∠POA = ∠AOS

∠1 = ∠8

Similarly,

∠2 = ∠3

∠4 = ∠5

∠6 = ∠7

∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360º

(∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360º

2∠1 + 2∠2 + 2∠5 + 2∠6 = 360º

2(∠1 + ∠2) + 2(∠5 + ∠6) = 360º

(∠1 + ∠2) + (∠5 + ∠6) = 180º

∠AOB + ∠COD = 180º

Similarly, we can prove that ∠BOC + ∠DOA = 180º

Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Question 12:

A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see given figure). Find the sides AB and AC.

 

Answer:

Let the given circle touch the sides AB and AC of the triangle at point E and F respectively and the length of the line segment AF be x.

In ΔABC,

CF = CD = 6cm (Tangents on the circle from point C)

BE = BD = 8cm (Tangents on the circle from point B)

AE = AF = x (Tangents on the circle from point A)

AB = AE + EB = x + 8

BC = BD + DC = 8 + 6 = 14

CA = CF + FA = 6 + x

2s = AB + BC + CA

= x + 8 + 14 + 6 + x

= 28 + 2x

s = 14 + x

Area of ΔABC = Area of ΔOBC + Area of ΔOCA + Area of ΔOAB

Either x+14 = 0 or x − 7 =0

Therefore, x = −14and 7

However, x = −14 is not possible as the length of the sides will be negative.

Therefore, x = 7

Hence, AB = x + 8 = 7 + 8 = 15 cm

CA = 6 + x = 6 + 7 = 13 cm

Question 11:

Prove that the parallelogram circumscribing a circle is a rhombus.

Answer:

Since ABCD is a parallelogram,

AB = CD …(1)

BC = AD …(2)

It can be observed that

DR = DS (Tangents on the circle from point D)

CR = CQ (Tangents on the circle from point C)

BP = BQ (Tangents on the circle from point B)

AP = AS (Tangents on the circle from point A)

Adding all these equations, we obtain

DR + CR + BP + AP = DS + CQ + BQ + AS

(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)

CD + AB = AD + BC

On putting the values of equations (1) and (2) in this equation, we obtain 2AB = 2BC

AB = BC …(3)

Comparing equations (1), (2), and (3), we obtain

AB = BC = CD = DA

Hence, ABCD is a rhombus.

Question 10:

Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Answer:

 

Let us consider a circle centered at point O. Let P be an external point from which two tangents PA and PB are drawn to the circle which are touching the circle at point A and B respectively and AB is the line segment, joining point of contacts A and B together such that it subtends ∠AOB at center O of the circle.

It can be observed that

OA (radius) ⊥ PA (tangent)

Therefore, ∠OAP = 90°

Similarly, OB (radius) ⊥ PB (tangent)

∠OBP = 90°

In quadrilateral OAPB,

Sum of all interior angles = 360º

∠OAP +∠APB+∠PBO +∠BOA = 360º

90º + ∠APB + 90º + ∠BOA = 360º

∠APB + ∠BOA = 180º

Hence, it can be observed that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.

Question 9:

In the given figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that ∠AOB=90º .

 

Answer:

Let us join point O to C.

In ΔOPA and ΔOCA,

OP = OC (Radii of the same circle)

AP = AC (Tangents from point A)

AO = AO (Common side)

ΔOPA ≅  ΔOCA (SSS congruence criterion)

Therefore, P ↔ C, A ↔ A, O ↔ O

∠POA = ∠COA … (i)

Similarly, ΔOQB  ≅  ΔOCB

∠QOB = ∠COB … (ii)

Since POQ is a diameter of the circle, it is a straight line.

Therefore, ∠POA + ∠COA + ∠COB + ∠QOB = 180º

From equations (i) and (ii), it can be observed that

2∠COA + 2 ∠COB = 180º

∠COA + ∠COB = 90º

∠AOB = 90°

 Question 8:

A quadrilateral ABCD is drawn to circumscribe a circle (see given figure) Prove that AB + CD = AD + BC

 Answer:

It can be observed that

DR = DS (Tangents on the circle from point D) … (1)

CR = CQ (Tangents on the circle from point C) … (2)

BP = BQ (Tangents on the circle from point B) … (3)

AP = AS (Tangents on the circle from point A) … (4)

Adding all these equations, we obtain

DR + CR + BP + AP = DS + CQ + BQ + AS

(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)

CD + AB = AD + BC

 Question 7:

Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Answer:

Let the two concentric circles be centered at point O. And let PQ be the chord of the larger circle which touches the smaller circle at point A. Therefore, PQ is tangent to the smaller circle.

OA ⊥ PQ (As OA is the radius of the circle)

Applying Pythagoras theorem in ΔOAP, we obtain

OA² + AP² = OP²

3² + AP² = 5²

9 + AP² = 25

AP² = 16

AP = 4

In ΔOPQ,

Since OA ⊥ PQ,

AP = AQ (Perpendicular from the center of the circle bisects the chord)

∴ PQ = 2AP = 2 × 4 = 8

Therefore, the length of the chord of the larger circle is 8 cm.

Question 6:

The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

Answer:

Let us consider a circle centered at point O.

AB is a tangent drawn on this circle from point A.

Given that,

OA = 5cm and AB = 4 cm

In ΔABO,

OB ⊥ AB (Radius ⊥ tangent at the point of contact)

Applying Pythagoras theorem in ΔABO, we obtain

AB² + BO² = OA²

4² + BO² = 5²

16 + BO² = 25

BO² = 9

BO = 3

Hence, the radius of the circle is 3 cm.

Question 5:

Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Answer:

Let us consider a circle with centre O. Let AB be a tangent which touches the circle at P.

We have to prove that the line perpendicular to AB at P passes through centre O. We shall prove this by contradiction method.

Let us assume that the perpendicular to AB at P does not pass through centre O. Let it pass through another point O’. Join OP and O’P.

As perpendicular to AB at P passes through O’, therefore,

∠O’PB = 90° … (1)

O is the centre of the circle and P is the point of contact. We know the line joining the centre and the point of contact to the tangent of the circle are perpendicular to each other.

∴ ∠OPB = 90° … (2)

Comparing equations (1) and (2), we obtain

∠O’PB = ∠OPB … (3)

From the figure, it can be observed that,

∠O’PB < ∠OPB … (4)

Therefore, ∠O’PB = ∠OPB is not possible. It is only possible, when the line O’P coincides with OP.

Therefore, the perpendicular to AB through P passes through centre O.

Question 4:

Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Answer:

 

Let AB be a diameter of the circle. Two tangents PQ and RS are drawn at points A and B respectively.

Radius drawn to these tangents will be perpendicular to the tangents.

Thus, OA ⊥ RS and OB ⊥ PQ

∠OAR = 90º

∠OAS = 90º

∠OBP = 90º

∠OBQ = 90º

It can be observed that

∠OAR = ∠OBQ (Alternate interior angles)

∠OAS = ∠OBP (Alternate interior angles)

Since alternate interior angles are equal, lines PQ and RS will be parallel.

Question 3:

If tangents PA and PB from a point P to a circle with centre O are inclined to each other an angle of 80º , then ∠POA is equal to

(A) 50º       (B) 60º        (C) 70º          (D) 80º

Answer:

It is given that PA and PB are tangents.

Therefore, the radius drawn to these tangents will be perpendicular to the tangents.

Thus, OA ⊥ PA and OB ⊥ PB

∠OBP = 90º

∠OAP = 90º

In AOBP,

Sum of all interior angles = 360º

∠OAP + ∠APB +∠PBO + ∠BOA = 360º

90º + 80º +90º + BOA = 360º

∠BOA = 100º

In ΔOPB and ΔOPA,

AP = BP (Tangents from a point)

OA = OB (Radii of the circle)

OP = OP (Common side)

Therefore, ΔOPB ≅ ΔOPA (SSS congruence criterion)

A ↔ B, P ↔ P, O ↔ O

And thus, ∠POB = ∠POA

∠POA = 1/2 ∠AOB = 100º/2 = 50º

Hence, alternative (A) is correct.

Question 2:

In the given figure, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110º , then ∠PTQ is equal to

(A) 60º         (B) 70º            (C) 80º             (D) 90º

Answer:

It is given that TP and TQ are tangents.

Therefore, radius drawn to these tangents will be perpendicular to the tangents.

Thus, OP ⊥ TP and OQ ⊥ TQ

∠OPT = 90º

∠OQT = 90º

In quadrilateral POQT,

Sum of all interior angles = 360º

∠OPT + ∠POQ +∠OQT + ∠PTQ = 360º

⇒ 90º + 110º + 90º + PTQ = 360º

⇒ PTQ = 70º

Hence, alternative (B) is correct.

Question 1:

From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is

(A) 7 cm       (B) 12 c        (C) 15 cm       (D) 24.5 cm

Answer:

Let O be the centre of the circle.

Given that,

OQ = 25cm and PQ = 24 cm

As the radius is perpendicular to the tangent at the point of contact,

Therefore, OP ⊥ PQ

Applying Pythagoras theorem in ΔOPQ, we obtain

OP² + PQ² = OQ²

OP² + 24² = 25²

OP² = 625 − 576

OP² = 49

OP = 7

Therefore, the radius of the circle is 7 cm.

Hence, alternative (A) is correct

Question 16:

The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m. from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

Answer:

 

Let AQ be the tower and R, S are the points 4m, 9m away from the base of the tower respectively.

The angles are complementary. Therefore, if one angle is θ, the other will be 90 − θ.

In ΔAQR,

 

In ΔAQS,

 

However, height cannot be negative.

Therefore, the height of the tower is 6 m.

Question 15:

A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car as an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Answer:

Let AB be the tower.

Initial position of the car is C, which changes to D after six seconds.

In ΔADB,

Time taken by the car to travel distance DC 

Time taken by the car to travel distance DB 

Question 14:

A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during the interval.

Answer:

Let the initial position A of balloon change to B after some time and CD be the girl.

In ΔACE,

Question 13:

As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Answer:

Let AB be the lighthouse and the two ships be at point C and D respectively.

Therefore, the distance between the two ships is 75(√3 – 1)m.

Question 12:

From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Answer:

Let AB be a building and CD be a cable tower.

In ΔABD,

Therefore, the height of the cable tower is 7(√3 + 1)m.